$\dfrac{ t - 5u }{ -4 } = \dfrac{ 5t - 8v }{ 9 }$ Solve for $t$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ t - 5u }{ -{4} } = \dfrac{ 5t - 8v }{ 9 }$ $-{4} \cdot \dfrac{ t - 5u }{ -{4} } = -{4} \cdot \dfrac{ 5t - 8v }{ 9 }$ $t - 5u = -{4} \cdot \dfrac { 5t - 8v }{ 9 }$ Multiply both sides by the right denominator. $t - 5u = -4 \cdot \dfrac{ 5t - 8v }{ {9} }$ ${9} \cdot \left( t - 5u \right) = {9} \cdot -4 \cdot \dfrac{ 5t - 8v }{ {9} }$ ${9} \cdot \left( t - 5u \right) = -4 \cdot \left( 5t - 8v \right)$ Distribute both sides ${9} \cdot \left( t - 5u \right) = -{4} \cdot \left( 5t - 8v \right)$ ${9}t - {45}u = -{20}t + {32}v$ Combine $t$ terms on the left. ${9t} - 45u = -{20t} + 32v$ ${29t} - 45u = 32v$ Move the $u$ term to the right. $29t - {45u} = 32v$ $29t = 32v + {45u}$ Isolate $t$ by dividing both sides by its coefficient. ${29}t = 32v + 45u$ $t = \dfrac{ 32v + 45u }{ {29} }$